Balls and Boxes
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 260 Accepted Submission(s): 187
Problem Description
Mr. Chopsticks is interested in random phenomena, and he conducts an experiment to study randomness. In the experiment, he throws n balls into m boxes in such a manner that each ball has equal probability of going to each boxes. After the experiment, he calculated the statistical variance V as
V=∑mi=1(Xi−X¯)2m
where Xi is the number of balls in the ith box, and X¯ is the average number of balls in a box.
Your task is to find out the expected value of V.
Input
The input contains multiple test cases. Each case contains two integers n and m (1 <= n, m <= 1000 000 000) in a line.
The input is terminated by n = m = 0.
Output
For each case, output the result as A/B in a line, where A/B should be an irreducible fraction. Let B=1 if the result is an integer.
Sample Input
2 1
2 2
0 0
Sample Output
0/1
1/2
Hint
In the second sample, there are four possible outcomes, two outcomes with V = 0 and two outcomes with V = 1.
Author
SYSU
Source
2016 Multi-University Training Contest 7
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题意：给你n个球,m个盒子，每个球落入每个盒子的概率是等可能的，求方差的期望值。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x7f7f7f7f
#define FOR(i,n) for(int i=1;i<=n;i++)
#define CT continue;
#define PF printf
#define SC scanf
const int mod=1000000007;
const int N=1e3+10;
ll gcd(ll a,ll b)
{
if(b==0) return a;
else return gcd(b,a%b);
}
int main()
{
ll n,m;
while(~scanf("%lld%lld",&n,&m)&&(n||m))
{
ll fenzi=n*(m-1),fenmu=m*m;
while(1)
{
ll k=gcd(fenzi,fenmu);
if(k==1) break;
fenzi/=k;fenmu/=k;
}
printf("%lld/%lld\n",fenzi,fenmu);
}
return 0;
}
分析：比赛时就感觉是个什么分布，，但是学的很多又忘了，最后百度了一下，才发现可以二项分布做；
对于每个盒子，每个球落入其中的概率是p=1/m；
那么总共n个球p(x=k)=C（n,k）*p^k*(1-p)^(n-k),显然的二项分布；
二项分布数学期望E(x)=np(n是实验次数，p是每次试验球落入盒子的概率)；
方差D（x）=np(1-p)
本题中D（x）=n/m*(1-1/m)=n*(m-1)/（m^2）;
然后因为每个盒子是平等的，方差又是描述数据的混乱程度，所以多个均等的盒子的方差与单个盒子方差
是一样的